The mid-points of the sides of a triangle are $(2, 1)$,$(-1, -3)$,and $(4, 5)$. The coordinates of its vertices are:

The perimeter of a triangle is $16 \text{ cm}$, one of the sides is of length $6 \text{ cm}$. If the area of the triangle is $12 \text{ cm}^2$, then the triangle is:

Let $ABC$ be a triangle such that the equations of lines $AB$ and $AC$ are $3y-x=2$ and $x+y=2$,respectively,and the points $B$ and $C$ lie on the $x$-axis. If $P$ is the orthocentre of the triangle $ABC$,then the area of the triangle $PBC$ is equal to

Two mutually perpendicular straight lines through the origin form an isosceles triangle with the line $2x + y = 5$. Then the area of the triangle is:

The length of the altitude through $A$ of the triangle $ABC$,where $A \equiv (-3, 0)$,$B \equiv (4, -1)$,and $C \equiv (5, 2)$,is

$A$ triangle $ABC$ lying in the first quadrant has two vertices as $A(1, 2)$ and $B(3, 1)$. If $\angle BAC = 90^{\circ}$ and $\text{ar}(\Delta ABC) = 5\sqrt{5}$ sq. units,then the abscissa of the vertex $C$ is